Find all integers n for which the equation x³ - 13x + n = 0 has three integer roots.

Let a, b and c be the three roots of the given equation. Without loss of generality, we can assume that |a| >= | b| > = |c|
From Vieta's formula for Polynomial roots, we obtain:
(I) a+b+c = 0 (II) ab+bc+ca = -13; (III) abc = -n
If c=0, then a=-b and ab =-13, giving:
 (a, b)=(sqrt(13), - sqrt(13)); (-sqrt(13), sqrt(13)) , which is a contradiction.
Now, from (I) and (II), we obtain:
 a^2 + b^2+c^2 = 26;
Clearly, all possible non-zero integer solutions of the above equation are:
(a,b,c) = (4,3,1);  (4,3,-1); (4,-3,1); (4,-3,-1); (-4,3,1); (-4,3,-1); (-4,-3,1); -(4,-3,-1).
Only, (a,b,c) = (4,-3,-1); (-4,3,1) satisfies both the conditions (I) and (II).
Now, (a,b,c) = (4,-3,-1) gives n = -abc = -12
and, (a,b,c) = (-4,3,1) gives n = -abc = 12.
Hence, the required values of n satisfying all the conditions of the puzzle are either  +12 or -12.

Edited on October 14, 2006, 10:41 pm

Posted by K Sengupta on 2006-10-14 22:39:25